Odpowiedzi

  • Roma
  • Community Manager
2010-03-12T23:03:19+01:00
1.
a = 2sin3α - tg3α
b = cos3α + cos4α
α =15°

a*b = (2sin3α - tg3α)*(cos3α + cos4α) dla α =15°
= (2sin45° - tg45°)*(cos45° + cos60°) = (2*√²/₂ - 1)*(√²/₂ + ½) = (√2 - 1)*(√²/₂ + ½) = ²/₂ + √²/₂ - √²/₂ - ½ = 1 - ½ = ½

2.
(sin405° * ctg150°) / (cos180° * tg210°) = [sin(45°+360°) * ctg(90°+60°)] / [- 1 * tg(180°+30°)] = [sin45° * (-tg60°)] /(-1 * tg30°) = [√²/₂ * (- √3)] / (-1 * √³/₃) = (- √²*√³/₂) : (- √³/₃) = (√²*√³/₂) * (³/√₃) = ³*√²/₂